

A249280


Repeatedly apply 'Reverse and add' to n. a(n) gives the number of steps needed to reach a sum containing each digit from 0 to 9 at least once.


1



38, 37, 35, 36, 54, 34, 45, 35, 48, 53, 52, 33, 51, 44, 32, 34, 50, 47, 43, 52, 33, 51, 44, 32, 34, 50, 47, 43, 42, 33, 51, 44, 32, 34, 50, 47, 43, 42, 31, 51, 44, 32, 34, 50, 47, 43, 42, 31, 41, 44, 32, 34, 50, 47, 43, 42, 31, 41, 33, 32, 34, 50, 47, 43, 42
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OFFSET

1,1


COMMENTS

Conjecture 1: a(n) exists for all n.
Conjecture 2: There exists an upper bound c such that a(n) < c for all n.
The conjectures seem highly likely, especially since a(n) = 0 for almost all n. A lower bound for c is a(1418993) = 73. (Checked to 10^9.)  Charles R Greathouse IV, Oct 28 2014


LINKS

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Felix Fröhlich, C++ program for this sequence


PROG

(PARI) fromdigits(v, b=10)=subst(Pol(v), 'x, b) \\ needed for gp < 2.63 or so
A056964(n)=fromdigits(Vecrev(digits(n)))+n
ispan(n)=#Set(digits(n))==10
a(n)=my(k); while(!ispan(n), n=A056964(n); k++); k \\ Charles R Greathouse IV, Oct 28 2014


CROSSREFS

Cf. A056964.
Sequence in context: A033974 A143721 A070725 * A022994 A023480 A036743
Adjacent sequences: A249277 A249278 A249279 * A249281 A249282 A249283


KEYWORD

nonn,base,easy


AUTHOR

Felix Fröhlich, Oct 26 2014


STATUS

approved



